Final answer:
The exponential generating function for the sequence 1, -1, 1, -1, 1, -1, ... is the real part of e^(ix), which is cos(x), represented as ∑((-1)^n x^(2n)/(2n)!) from n=0 to infinity.
Step-by-step explanation:
To find the exponential generating function (EGF) for the sequence 1, -1, 1, -1, 1, -1, ..., we can consider that this pattern represents the coefficients of the power series expansion of some function. Since the coefficients alternate between 1 and -1, one possible function that has this property is the cosine function. By comparing it to the Maclaurin series expansion for cosine, we have cos(x) = 1 - x2/2! + x4/4! - ..., which only contains the even powers of x. However, in our case, we want coefficients for all degrees of x, not just the evens. To achieve this, we realize that eix = cos(x) + i sin(x) has the desired alternating pattern in the real part of its Maclaurin series. Thus, the EGF we are looking for is the real part of eix, which is Re(eix). Hence, the EGF for the given sequence is cos(x), which can be written as ∑((-1)n x2n/(2n)!) from n=0 to infinity, representing the real part of eix.